A) \[-0.6\]
B) \[-0.88\]
C) 0.88
D) 0.6
Correct Answer: D
Solution :
\[{{E}^{o}}_{cell}={{E}_{cathode}}-{{E}_{anode}}\] \[{{E}_{cathode(S{{n}^{2+}}/Sn)}}=-0.14\] \[{{E}_{anode(Z{{n}^{2+}}/Zn)}}=-0.74\] \[{{E}^{o}}_{cell}=-0.14-(-0.74)\] \[=0.74-0.14\] \[=0.6\,\text{V}\]You need to login to perform this action.
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