Avoltaic 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.

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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Explanation:

We must use the Nernst equation

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

the answer is c i believe

that's what i got you can count the number of atoms