To calculate the standard cell potential for the reaction: Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s), we need to identify the half-reactions and use the given standard reduction potentials.

Step 1: Identify the half-reactions
The reaction involves:
- Reduction: Ag⁺(aq) + e^- → Ag(s) with E⁰ = x V
- Oxidation: Fe²⁺(aq) → Fe³⁺(aq) + e^- (This is not directly given; we need to relate it to the provided potentials)

Step 2: Relate the oxidation half-reaction to given potentials
The oxidation half-reaction is the reverse of the reduction: Fe³⁺(aq) + e^- → Fe²⁺(aq). The standard reduction potential for this is given as E⁰_{Fe³⁺/Fe²⁺} = z V. Therefore, for the oxidation (reverse reaction), the potential is -z V.

Step 3: Calculate the standard cell potential
The standard cell potential E⁰_cell is calculated as:
$E_{\mathrm{cell}}^0=E_{\mathrm{reduction}}^0-E_{\mathrm{oxidation}}^0=E_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^0-E_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}^0=x-z$ V

Final Answer: The standard cell potential is x - z V. Among the options, this corresponds to "x – z".

Related Topics and Formulae

Standard Cell Potential Calculation: The standard cell potential for a galvanic cell is given by E⁰_cell = E⁰_cathode - E⁰_anode, where cathode is the reduction half-cell and anode is the oxidation half-cell. Always ensure the half-reactions are written as reductions, and reverse the sign for oxidation.

Nernst Equation: For non-standard conditions, the cell potential can be calculated using the Nernst equation: $E=E_{\mathrm{cell}}^0-\frac{0.059}{n}logQ$ at 298 K, where n is the number of electrons transferred and Q is the reaction quotient.