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 E_cell if E⁰_cell = 3.17 V.

The half cell reactions can be written as:

A n o d e : M g_{(s)} \to M g_{(0.130 M)}^{2 +} + 2 e^{-}

C a t h o d e : 2 A g_{(0.0001 M)}^{+} + 2 e^{-} \to 2 A g_{(s)}

The cell can be represented as:

M g_{(s)} / M g_{(0.130 M)}^{2 +} | | A g_{(0.0001 M)}^{+} / A g_{(s)}

Calculation of E_cell:

N u m b e r o f e l e c t r o n s e x c h a n g e d, n = 2

R e a c t i o n Q u o t i e n t, Q = \frac{[M g^{2 +}]}{[A g^{+}]^{2}}

Q = \frac{0.130}{(1 \times 10^{- 4})^{2}} = 1.3 \times 10^{7}

E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{n} l o g Q

E_{c e l l} = 3.17 - \frac{0.0591}{2} l o g (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: E⁰_cell = 1.05 V.

The half cell reactions can be written as:

A n o d e : N i_{(s)} \to N i_{(0.160 M)}^{2 +} + 2 e^{-}

C a t h o d e : 2 A g_{(0.002 M)}^{+} + 2 e^{-} \to 2 A g_{(s)}

The cell can be represented as:

N i_{(s)} / N i_{(0.130 M)}^{2 +} | | A g_{(0.002 M)}^{+} / A g_{(s)}

Calculation of E_cell:

N u m b e r o f e l e c t r o n s e x c h a n g e d, n = 2

R e a c t i o n Q u o t i e n t, Q = \frac{[N i^{2 +}]}{[A g^{+}]^{2}}

Q = \frac{0.160}{(0.002)^{2}} = 4 \times 10^{4}

E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{n} l o g Q

E_{c e l l} = 1.05 - \frac{0.0591}{2} l o g (4 \times 10^{4}) = 0.91 V

Calculate the equilibrium constant of the reaction:
Cu_(s) + 2Ag⁺_(aq) → Cu²⁺_(aq) + 2Ag_(s)
Given that: E⁰_cell = 0.46 V.

The half cell reactions can be written as:

A n o d e : C u_{(s)} \to C u_{(a q)}^{2 +} + 2 e^{-}

C a t h o d e : 2 A g_{(a q)}^{+} + 2 e^{-} \to 2 A g_{(s)}

N u m b e r o f e l e c t r o n s e x c h a n g e d, n = 2

At Equilibrium:

E_{c e l l}^{0} = \frac{0.0591}{n} l o g K_{c}

0.46 = \frac{0.0591}{2} l o g K_{c}

K_{c} = 3.92 \times 10^{15}

The cell in which the following reaction occurs:
2Fe³⁺_(aq) + 2I^-_(aq) → 2Fe⁺_(aq) + I_2(aq)
has E⁰_cell = 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:

A n o d e : 2 I_{(a q)}^{-} \to I_{2 (s)} + 2 e^{-}

C a t h o d e : 2 F e_{(a q)}^{3 +} + 2 e^{-} \to 2 F e_{(a q)}^{2 +}

N u m b e r o f e l e c t r o n s e x c h a n g e d, n = 2

Calculation of Standard Gibbs energy:

Δ G^{0} = - n F E_{c e l l}^{0}

Δ G^{0} = - 2 \times 96500 \times 0.236

K_{c} = - 45548 J

Calculation of equilibrium constant:

E_{c e l l}^{0} = \frac{0.0591}{n} l o g K_{c}

0.236 = \frac{0.0591}{2} l o g K_{c}

l o g K_{c} = \frac{2 \times 0.236 \times 10}{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:

Δ G^{0} = - n F E_{c e l l}^{0}

Δ G^{0} = - 2 \times 96500 \times 1.1

Δ G^{0} = - 212300 J = 212.3 K J

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

We know that pH can be calculated as:

p H = - l o g [H^{+}]

10 = - l o g [H^{+}]

[H^{+}] = 10^{- 10} M

For hydrogen electrode:

H^{+} + e^{-} \to \frac{1}{2} H_{2}

Q = \frac{1}{[H^{+}]} = \frac{1}{10^{- 10}} = 10^{10}

Number of electrons exchanges, n = 1
Applying Nernst Equation:

E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{n} l o g Q

E_{c e l l} = 0 - \frac{0.0591}{1} l o g (10^{10}) = - 0.591 V