The plot of Reduction Potential $(E_{r e d}) v s \log [A g^{+}]$ for $A g^{+} | A g$ half cell is given as

Given:
$E_{A g^{+} | A g}^{0} = 0.8 V$ (independent of temperature) ; $\frac{2.303 R T}{F} = 0.06 a t 300 K; t a n θ = 0.062$
What is the temperature of the solution in $° C$

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 37.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$A g^{+} (a q) + e^{-} \overset{}{⇌} A g (s)$

$E_{r e d} = E^{0} - \frac{R T}{n F} \log \frac{1}{[A g^{+}]}$

$\Rightarrow E_{r e d} = E^{0} + \frac{R T}{F} \log [A g^{+}]$

$∴ S l o p e o f E_{r e d} v s \log [A g^{+}], M = \frac{R T}{F}$

$G i v e n \frac{R T}{F} = 0.06 a t 300 K a n d 0.0062 a t t h e$ temperature of the solution.

$\frac{\frac{R T_{1}}{F}}{\frac{R T_{2}}{F}} = \frac{0.06}{0.062}$

$T_{2} = \frac{0.062}{0.06} \times T_{1} = \frac{0.062}{0.06} \times 300 = 310 K = 37 ° C$

Watch 3-min video & get full concept clarity