In a galvanic cell the following reaction takes place at `298^(@)K`
`Cr_(2)O_(7)^(2-)+14H^(+)+6Fe^(2+)to2Cr^(3+)+6Fe^(3+)+7H_(2)O` given that
`:E^(o)(Cr_(2)O_(7)^(2-),H^(+),Cr^(3+)//Pt)=1.33V`
`E^(o)(Fe^(3+),Fe^(2+)//Pt)=0.77V`. The standard e.m.f. of the cell is

Options

A. (1.33+0.77)V
B. (1.33-0.77)V
C. `-(1.33+0.77)V`
D. `(-1.33+0.77)V`

Answer

To find the standard electromotive force (e.m.f.) of the galvanic cell for the given reaction, we can follow these steps:

### Step 1: Identify the half-reactions
The overall reaction is:
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O} \]

From this reaction, we can identify the half-reactions:
1. Reduction half-reaction (at the cathode):
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]
with \( E^\circ = 1.33 \, \text{V} \)

2. Oxidation half-reaction (at the anode):
\[ 6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^- \]
with \( E^\circ = 0.77 \, \text{V} \)

### Step 2: Determine the standard e.m.f. of the cell
The standard e.m.f. of the cell can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
- \( E^\circ_{\text{cathode}} = 1.33 \, \text{V} \)
- \( E^\circ_{\text{anode}} = 0.77 \, \text{V} \)

Thus,
\[ E^\circ_{\text{cell}} = 1.33 \, \text{V} - 0.77 \, \text{V} \]

### Step 3: Calculate the result
Perform the subtraction:
\[ E^\circ_{\text{cell}} = 1.33 - 0.77 = 0.56 \, \text{V} \]

### Final Answer
The standard e.m.f. of the cell is:
\[ E^\circ_{\text{cell}} = 0.56 \, \text{V} \]

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