Possible Cathode Reactions (Reduction)

Materials Science · FE Reference Handbook section

Materials Science

0 formulas

10 exam-style examples

~45 min

All Materials Science lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

When dissimilar metals are in contact, the more electropositive one becomes the anode in a corrosion cell. Different regions of
carbon steel can also result in a corrosion reaction: e.g., cold-worked regions are anodic to noncold-worked; different oxygen
concentrations can cause oxygen-deficient regions to become cathodic to oxygen-rich regions; grain boundary regions are
anodic to bulk grain; in multiphase alloys, various phases may not have the same galvanic potential.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Resistivity of a conductor and capacitance of a parallel plate — Possible Cathode Reactions (Reduction)

A conductor of resistivity 0.00e+0 Ω·m is 1.5 m long with a cross-section of 1.8e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 5.5), plate area 0.0140 m² and spacing 0.00080 m is charged to 78 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 1.5 m, A = 1.8e-5 m^{2}$
$\varepsilon_r = 5.5, A_p = 0.0140 m^{2}, d = 0.00080 m$
$V = 78 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 0.00e+0(1.5)/1.8e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(5.5)(0.0140)/0.00080 = 8.522e-10 F$
Formula
$Q = C V$
Substituting
$Q = 8.522e-10(78) = 6.647e-8 C$

Answer:

R = 0.000 \Omega, C = 8.52e-10 F, Q = 6.65e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.55e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

Example 2

Resistivity of a conductor and capacitance of a parallel plate — Possible Cathode Reactions (Reduction) (2)

A conductor of resistivity 0.00e+0 Ω·m is 2.5 m long with a cross-section of 1.7e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 7.0), plate area 0.0010 m² and spacing 0.00020 m is charged to 62 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 2.5 m, A = 1.7e-5 m^{2}$
$\varepsilon_r = 7.0, A_p = 0.0010 m^{2}, d = 0.00020 m$
$V = 62 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 0.00e+0(2.5)/1.7e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(7.0)(0.0010)/0.00020 = 3.099e-10 F$
Formula
$Q = C V$
Substituting
$Q = 3.099e-10(62) = 1.921e-8 C$

Answer:

R = 0.000 \Omega, C = 3.10e-10 F, Q = 1.92e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 4.43e-11 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

Example 3

Resistivity of a conductor and capacitance of a parallel plate — Possible Cathode Reactions (Reduction) (3)

A conductor of resistivity 0.00e+0 Ω·m is 0.5 m long with a cross-section of 9.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 6.0), plate area 0.0050 m² and spacing 0.00100 m is charged to 92 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 0.5 m, A = 9.0e-6 m^{2}$
$\varepsilon_r = 6.0, A_p = 0.0050 m^{2}, d = 0.00100 m$
$V = 92 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 0.00e+0(0.5)/9.0e-6 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(6.0)(0.0050)/0.00100 = 2.656e-10 F$
Formula
$Q = C V$
Substituting
$Q = 2.656e-10(92) = 2.444e-8 C$

Answer:

R = 0.000 \Omega, C = 2.66e-10 F, Q = 2.44e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 4.43e-11 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

Example 4

Resistivity of a conductor and capacitance of a parallel plate — Possible Cathode Reactions (Reduction) (4)

A conductor of resistivity 0.00e+0 Ω·m is 1.0 m long with a cross-section of 5.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 5.5), plate area 0.0130 m² and spacing 0.00040 m is charged to 169 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 1.0 m, A = 5.0e-6 m^{2}$
$\varepsilon_r = 5.5, A_p = 0.0130 m^{2}, d = 0.00040 m$
$V = 169 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 0.00e+0(1.0)/5.0e-6 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(5.5)(0.0130)/0.00040 = 1.583e-9 F$
Formula
$Q = C V$
Substituting
$Q = 1.583e-9(169) = 2.675e-7 C$

Answer:

R = 0.000 \Omega, C = 1.58e-9 F, Q = 2.67e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 2.88e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

Example 5

Resistivity of a conductor and capacitance of a parallel plate — Possible Cathode Reactions (Reduction) (5)

A conductor of resistivity 0.00e+0 Ω·m is 1.5 m long with a cross-section of 2.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 4.5), plate area 0.0060 m² and spacing 0.00020 m is charged to 102 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 1.5 m, A = 2.0e-6 m^{2}$
$\varepsilon_r = 4.5, A_p = 0.0060 m^{2}, d = 0.00020 m$
$V = 102 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 0.00e+0(1.5)/2.0e-6 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(4.5)(0.0060)/0.00020 = 1.195e-9 F$
Formula
$Q = C V$
Substituting
$Q = 1.195e-9(102) = 1.219e-7 C$

Answer:

R = 0.000 \Omega, C = 1.20e-9 F, Q = 1.22e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 2.66e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

Example 6

Resistivity of a conductor and capacitance of a parallel plate — Possible Cathode Reactions (Reduction) (6)

A conductor of resistivity 0.00e+0 Ω·m is 1.5 m long with a cross-section of 1.9e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 6.0), plate area 0.0040 m² and spacing 0.00070 m is charged to 8 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 1.5 m, A = 1.9e-5 m^{2}$
$\varepsilon_r = 6.0, A_p = 0.0040 m^{2}, d = 0.00070 m$
$V = 8 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 0.00e+0(1.5)/1.9e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(6.0)(0.0040)/0.00070 = 3.036e-10 F$
Formula
$Q = C V$
Substituting
$Q = 3.036e-10(8) = 2.429e-9 C$

Answer:

R = 0.000 \Omega, C = 3.04e-10 F, Q = 2.43e-9 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 5.06e-11 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

Example 7

Resistivity of a conductor and capacitance of a parallel plate — Possible Cathode Reactions (Reduction) (7)

A conductor of resistivity 1.00e-6 Ω·m is 4.0 m long with a cross-section of 1.7e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 2.0), plate area 0.0080 m² and spacing 0.00040 m is charged to 185 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 4.0 m, A = 1.7e-5 m^{2}$
$\varepsilon_r = 2.0, A_p = 0.0080 m^{2}, d = 0.00040 m$
$V = 185 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 1.00e-6(4.0)/1.7e-5 = 0.2353 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(2.0)(0.0080)/0.00040 = 3.542e-10 F$
Formula
$Q = C V$
Substituting
$Q = 3.542e-10(185) = 6.552e-8 C$

Answer:

R = 0.235 \Omega, C = 3.54e-10 F, Q = 6.55e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.77e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

Example 8

Resistivity of a conductor and capacitance of a parallel plate — Possible Cathode Reactions (Reduction) (8)

A conductor of resistivity 1.00e-6 Ω·m is 1.0 m long with a cross-section of 2.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 4.0), plate area 0.0100 m² and spacing 0.00010 m is charged to 72 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 1.0 m, A = 2.0e-6 m^{2}$
$\varepsilon_r = 4.0, A_p = 0.0100 m^{2}, d = 0.00010 m$
$V = 72 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 1.00e-6(1.0)/2.0e-6 = 0.5000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(4.0)(0.0100)/0.00010 = 3.542e-9 F$
Formula
$Q = C V$
Substituting
$Q = 3.542e-9(72) = 2.550e-7 C$

Answer:

R = 0.500 \Omega, C = 3.54e-9 F, Q = 2.55e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 8.85e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

Example 9

Resistivity of a conductor and capacitance of a parallel plate — Possible Cathode Reactions (Reduction) (9)

A conductor of resistivity 1.00e-6 Ω·m is 1.0 m long with a cross-section of 1.8e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 2.5), plate area 0.0040 m² and spacing 0.00050 m is charged to 93 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 1.0 m, A = 1.8e-5 m^{2}$
$\varepsilon_r = 2.5, A_p = 0.0040 m^{2}, d = 0.00050 m$
$V = 93 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 1.00e-6(1.0)/1.8e-5 = 0.0556 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(2.5)(0.0040)/0.00050 = 1.771e-10 F$
Formula
$Q = C V$
Substituting
$Q = 1.771e-10(93) = 1.647e-8 C$

Answer:

R = 0.056 \Omega, C = 1.77e-10 F, Q = 1.65e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 7.08e-11 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)

Example 10

Resistivity of a conductor and capacitance of a parallel plate — Possible Cathode Reactions (Reduction) (10)

A conductor of resistivity 0.00e+0 Ω·m is 0.5 m long with a cross-section of 1.2e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 8.0), plate area 0.0030 m² and spacing 0.00020 m is charged to 199 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 0.5 m, A = 1.2e-5 m^{2}$
$\varepsilon_r = 8.0, A_p = 0.0030 m^{2}, d = 0.00020 m$
$V = 199 V$

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = \rho L / A$
Substituting
$R = 0.00e+0(0.5)/1.2e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(8.0)(0.0030)/0.00020 = 1.062e-9 F$
Formula
$Q = C V$
Substituting
$Q = 1.062e-9(199) = 2.114e-7 C$

Answer:

R = 0.000 \Omega, C = 1.06e-9 F, Q = 2.11e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.33e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Possible Cathode Reactions (Reduction)