Question

Subject : Chemistry Class : XII Chapter : Electrochemistry

(a) Calculate the $emf$ for the given cell at 25 °C:



$\bold{\mathrm{Cr~|~Cr^{3+}\:\left(0.1\:M\right)~||~Fe^{2+}\:\left(0.01\:M\right)~|~Fe}}$
$\left[\mathrm{Given:~E^\circ_{Cr^{3+}}=-0.74\:V,~E^\circ_{Fe^{2+}/Fe}=-0.44\:V}\right]$

Answer 1

Answer:

Answer is in attachment.

Answer 2

According to Nernst equation :

Ecell=E∘cell−0.0591Vnlog[Zn2+][H+]2

Ecell=(0.76 V)−(0.0591 V)2log10−−3(10−2)2

=0.76 V −(0.02955 V ) log 10=(0.76−0.2955) V =0.7305 V