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ChemistryEdu Logo Electrochemistry | Questions Based on Nernst Equation#

Questions#

Represent the cell in which the following reaction takes place:
Mg(s) + 2Ag+(0.0001 M) → Mg2+(0.130 M) + 2Ag(s)
Calculate its Ecell if E0cell = 3.17 V.

  • The half cell reactions can be written as:
\[Anode:\ Mg_{(s)} → Mg^{2+}_{(0.130M)} + 2e^-\]
\[Cathode:\ 2Ag_{(0.0001\ M)}^+ + 2e^- → 2Ag_{(s)}\]
  • The cell can be represented as:
\[Mg_{(s)}/Mg^{2+}_{(0.130\ M)} || Ag^+_{(0.0001\ M)}/Ag_{(s)}\]
  • Calculation of Ecell:
\[Number\ of\ electrons\ exchanged,\ n = 2\]
\[Reaction\ Quotient,\ Q = {[Mg^{2+}] \over [Ag^+]^2}\]
\[Q = {0.130 \over (1 \times 10^{-4})^2} = 1.3 \times 10^7\]
\[E_{cell} = E^0_{cell} - {0.0591 \over n} logQ\]
\[E_{cell} = 3.17 - {0.0591 \over 2}log(1.3 \times 10^7) = 2.96\ V\]

Calculate the emf of cell in which the following reaction takes place:
Ni(s) + 2Ag+(0.002 M) → Ni2+(0.160 M) + 2Ag(s)
Given that: E0cell = 1.05 V.

  • The half cell reactions can be written as:
\[Anode:\ Ni_{(s)} → Ni^{2+}_{(0.160M)} + 2e^-\]
\[Cathode:\ 2Ag^+_{(0.002\ M)} + 2e^- → 2Ag_{(s)}\]
  • The cell can be represented as:
\[Ni_{(s)}/Ni^{2+}_{(0.130\ M)} || Ag^+_{(0.002\ M)}/Ag_{(s)}\]
  • Calculation of Ecell:
\[Number\ of\ electrons\ exchanged,\ n = 2\]
\[Reaction\ Quotient,\ Q = {[Ni^{2+}] \over [Ag^+]^2}\]
\[Q = {0.160 \over (0.002)^2} = 4 \times 10^4\]
\[E_{cell} = E^0_{cell} - {0.0591 \over n} logQ\]
\[E_{cell} = 1.05 - {0.0591 \over 2}log(4 \times 10^4) = 0.91\ V\]

Calculate the equilibrium constant of the reaction:
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
Given that: E0cell = 0.46 V.

  • The half cell reactions can be written as:
\[Anode:\ Cu_{(s)} → Cu^{2+}_{(aq)} + 2e^-\]
\[Cathode:\ 2Ag^+_{(aq)} + 2e^- → 2Ag_{(s)}\]
\[Number\ of\ electrons\ exchanged,\ n = 2\]
  • At Equilibrium:
\[E^0_{cell} = {0.0591 \over n}logK_c\]
\[0.46 = {0.0591 \over 2}logK_c\]
\[K_c = 3.92 \times 10^{15}\]

The cell in which the following reaction occurs:
2Fe3+(aq) + 2I-(aq) → 2Fe+(aq) + I2(aq)
has E0cell = 0.236 V$ at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the reaction.

  • The half cell reactions can be written as:
\[Anode:\ 2I^-_{(aq)} → I_{2(s)} + 2e^-\]
\[Cathode:\ 2Fe^{3+}_{(aq)} + 2e^- → 2Fe^{2+}_{(aq)}\]
\[Number\ of\ electrons\ exchanged,\ n = 2\]
  • Calculation of Standard Gibbs energy:
\[ΔG^0 = -nFE^0_{cell}\]
\[ΔG^0 = -2 \times 96500 \times 0.236\]
\[K_c = -45548\ J\]
  • Calculation of equilibrium constant:
\[E^0_{cell} = {0.0591 \over n}logK_c\]
\[0.236 = {0.0591 \over 2}logK_c\]
\[logK_c = {2 \times 0.236 \times 10 \over 0.0591}\]
\[K_c = 9.6 \times 10^7\]

The standard electrode potential for Daniell cell is 1.1 V. Calculate the standard Gibbs Energy for the reaction:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Number of electrons exchanged, n = 2

Standard Gibbs Energy of the reaction:

\[ΔG^0 = -nFE^0_{cell}\]
\[ΔG^0 = -2 \times 96500 \times 1.1\]
\[ΔG^0 = -212300\ J = 212.3\ KJ\]

Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

  • We know that pH can be calculated as:
\[pH = -log[H^+]\]
\[10 = -log[H^+]\]
\[[H^+] = 10^{-10} M\]
  • For hydrogen electrode:
\[H^+ + e^- → {1 \over 2} H_2\]
\[Q = {1 \over [H^+]} = {1 \over 10^{-10}} = 10^{10}\]

  • Number of electrons exchanges, n = 1

  • Applying Nernst Equation:

\[E_{cell} = E^0_{cell} - {0.0591 \over n}logQ\]
\[E_{cell} = 0 - {0.0591 \over 1}log(10^{10}) = -0.591\ V\]