Question

A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction is Ag (s)C AgCl (s) (aq) + e F= +0.222 V The concentrations of chloride ion in the two compartments are 0.0100 M and 1.55 M, respectively. At 25 °C, the cell emf is V.

Answer 1

Answer:

$\boxed{\text{0.130 V}}$

Explanation:

We must use the Nernst equation

$E = E^{\circ} - \dfrac{RT}{zF} \ln Q$

1. Write the equation for the cell reaction

The cell reaction is

                                                                                         E°/V  

  Anode: Ag(s) + Cl⁻(1.55 mol·L⁻¹) ⇌ AgCl(s) + e⁻;       +0.222

Cathode: AgCl(s) + e⁻ ⇌ Ag(s) + Cl⁻(0.0100 mol·L⁻¹);  -0.222

 Overall: Cl⁻(1.55 mol·L⁻¹) ⇌ Cl⁻(0.0100 mol·L⁻¹)          0.000

Step 2. Calculate E

(a) Data

  E° = 0 V

  R = 8.314 J·K⁻¹mol⁻¹

  T = 25 °C

  n = 1

  F = 96 485 C/mol

[Cl⁻]_ prod = 0.0100 mol·L⁻¹

[Cl⁻]_ react = 1.55      mol·L⁻¹

(b) Calculations:  

T = 25 + 273.15 = 298.15 K

$Q = \dfrac{\text{[Cl}^{-}]_{\text{prod}}}{\text{[Cl}^{-}]_{\text{react}}} = \dfrac{0.0100}{1.55} =0.006452\\\\E = 0 - \left (\dfrac{8.314 \times 298.15 }{1 \times 96485}\right ) \ln(0.006452)\\\\= -0.02569 \times (-5.043) = \textbf{0.130 V}\\\text{The cell potential is }\boxed{\textbf{0.130 V}}$

Answer 2

The cell voltage as written is 0.13 V.

The equation of the reaction is; Ag(s) + Cl(aq) -----> AgCl(s) (aq) + e- = +0.222 V

Since  the cathode and anode are the same, E°cell = 0.00 V

We also know that;

Q =  [0.0100 M]/[ 1.55 M]

= 0.00645

From the Nernst equation;

Ecell = E°cell - 0.0592/n log Q

Where n = 1

Ecell = 0.00 - 0.0592/1 log (0.00645)

Ecell = 0.13 V

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