At `20^(@)` C, the standard oxidation potential of Zn and Ag in water are:
`Zn(s)rarr Zn^(2+) (aq) + 2e^(-), E^(o) = 0.76V`,
`Ag(s) rarr Ag^(+)(aq) + E^(-) , E^(o) = - 0.80 V`
The standard EMF of the given reaction is:
`Zn+2Ag^(+)rarr 2AG+Zn^(+2)`

Options

A. `-1.56`V
B. `+1.56`V
C. `-0.04`V
D. `0.04` V

Answer

To solve the problem of finding the standard EMF of the given reaction:

**Given Reaction:**
\[ \text{Zn} + 2\text{Ag}^+ \rightarrow 2\text{Ag} + \text{Zn}^{2+} \]

**Step 1: Identify Oxidation and Reduction**
- In this reaction, zinc (Zn) is oxidized to zinc ions (\( \text{Zn}^{2+} \)), and silver ions (\( \text{Ag}^+ \)) are reduced to silver (Ag).
- Oxidation half-reaction:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]
- Reduction half-reaction:
\[ 2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag} \]

**Step 2: Write the Standard Electrode Potentials**
- The standard oxidation potential for zinc is given as:
\[ E^\circ (\text{Zn}^{2+}/\text{Zn}) = -0.76 \, \text{V} \]
- The standard reduction potential for silver is given as:
\[ E^\circ (\text{Ag}^+/Ag) = +0.80 \, \text{V} \]

**Step 3: Calculate the Standard EMF of the Cell**
- The standard EMF (\( E^\circ_{\text{cell}} \)) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
- Here, silver is the cathode (reduction occurs) and zinc is the anode (oxidation occurs):
\[ E^\circ_{\text{cell}} = E^\circ (\text{Ag}^+/Ag) - E^\circ (\text{Zn}^{2+}/\text{Zn}) \]
\[ E^\circ_{\text{cell}} = (+0.80 \, \text{V}) - (-0.76 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = 0.80 \, \text{V} + 0.76 \, \text{V} \]
\[ E^\circ_{\text{cell}} = 1.56 \, \text{V} \]

**Final Answer:**
The standard EMF of the given reaction is \( 1.56 \, \text{V} \).

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