`Ag^(+) + e^(-) rarr Ag, E^(0) = + 0.8 V and Zn^(2+) + 2e^(-) rarr Zn, E^(0) = -076 V`. Calculate the cell potential for the reaction, `2Ag + Zn^(2+) rarr Zn + 2Ag^(+)`

Answer

To calculate the cell potential for the reaction \(2Ag + Zn^{2+} \rightarrow Zn + 2Ag^{+}\), we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials (E°)
The half-reactions given in the question are:
1. \(Ag^{+} + e^{-} \rightarrow Ag\), with \(E^{0} = +0.80 \, V\)
2. \(Zn^{2+} + 2e^{-} \rightarrow Zn\), with \(E^{0} = -0.76 \, V\)

### Step 2: Determine the anode and cathode
- The half-reaction with the higher standard reduction potential will act as the cathode (reduction occurs here).
- The half-reaction with the lower standard reduction potential will act as the anode (oxidation occurs here).

In this case:
- \(Ag^{+} + e^{-} \rightarrow Ag\) (cathode, \(E^{0} = +0.80 \, V\))
- \(Zn \rightarrow Zn^{2+} + 2e^{-}\) (anode, \(E^{0} = -0.76 \, V\))

### Step 3: Write the oxidation half-reaction
The oxidation half-reaction for zinc is the reverse of its reduction:
\[ Zn \rightarrow Zn^{2+} + 2e^{-} \]

### Step 4: Calculate the cell potential (E°cell)
The cell potential can be calculated using the formula:
\[ E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode} \]
Substituting the values:
\[ E^{0}_{cell} = E^{0}_{Ag^{+}/Ag} - E^{0}_{Zn^{2+}/Zn} \]
\[ E^{0}_{cell} = 0.80 \, V - (-0.76 \, V) \]
\[ E^{0}_{cell} = 0.80 \, V + 0.76 \, V \]
\[ E^{0}_{cell} = 1.56 \, V \]

### Final Answer:
The cell potential for the reaction \(2Ag + Zn^{2+} \rightarrow Zn + 2Ag^{+}\) is \(E^{0}_{cell} = 1.56 \, V\).

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