Question

Calculate the emf of the following cell at 298 K :

Cr(s) / Cr3+ (0.1M) // Fe2+ (0.01M) / Fe(s)

[Given :

0 ECell

= + 0.30 V]

Answer 1

Sorry there's a mistake....
Here n=6.... Not 3...
So the calculation will be..
0.30-(0.059/6)log[(0.1)²/(0.01)³]
=>0.26V(ans)

Answer 2

Answer: The EMF of the cell is 0.290 V.

Explanation:

We are given:

$E^o_{cell}=+0.30V$

The given cell is:

$Cr(s)/Cr^{3+}(0.1M)||Fe^{2+}(0.01M)/Fe(s)$

Half reactions for the given cell follows:

Oxidation half reaction:  $Cr(s)\rightarrow Cr^{3+}(0.1M)+3e^-$       ( × 3)

Reduction half reaction:  $Fe^{2+}(0.01M)+2e^-\rightarrow Fe(s)$       ( × 2)

Net reaction:  $Cr(s)+Fe^{2+}(0.01M)\rightarrow Cr^{3+}(0.1M)+Fe(s)$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Cr^{3+}]}{[Fe^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = ?V

$E^o_{cell}$ = standard electrode potential of the cell = +0.30 V

n = number of electrons exchanged = 6

$[Cr^{3+}]=0.1M$

$[Fe^{2+}]=0.01M$

Putting values in above equation, we get:

$E_{cell}=0.30-\frac{0.059}{6}\times \log(\frac{0.1}{0.01})\\\\E_{cell}=0.290V$

Hence, the EMF of the cell is 0.290 V.