A) + 0.89 V
B) + 0.18 V
C) + 1.83 V
D) + 1.199 V
Correct Answer: A
Solution :
\[E{{_{Sn}^{o}}_{^{4+}/S{{n}^{2+}}}}=+0.15\,\text{V}\] \[E{{_{Cr}^{o}}_{^{3+}/Cr}}=-\,0.74\,\text{V}\] \[E_{cell}^{o}={{E}^{o}}_{cathode(RP)}-{{E}^{o}}_{anode(RP)}\] \[=0.15-(-0.74)\] \[=+\,0.89\,V.\]You need to login to perform this action.
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