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Question

Chemistry Question on Electrochemistry

E=cr3+/cr0=0.72V,Efe2+/fe0=0.42 E=^0_{cr^{3+}/cr}=-0.72 V, E^0_{fe^{2+}/fe}=-0.42 The potential for the cell Cr|Cr3+(0.1 M) || Fe2+(o.01 M) Fe is

A

-0.339 V

B

-0.26 V

C

0.26 V

D

0.3 V

Answer

0.26 V

Explanation

Solution

The correct option is (C): 0.26 V