A) 0.04736 V
B) 0.4736 V
C) 0.4736 mV
D) 0.04736 mV
Correct Answer: B
Solution :
\[{{E}^{o}}_{cell}=\frac{2.303RT}{nF}\log {{K}_{c}}\] \[=\frac{0.059}{2}\log (10\times {{10}^{15}})=\frac{0.059\times 16}{2}\]\[=0.472V\]You need to login to perform this action.
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