A) +1.83 V
B) +1.19 V
C) +0.18 V
D) +0.89 V
Correct Answer: D
Solution :
\[S{{n}^{4+}}+2e\xrightarrow{{}}S{{n}^{2+}};E_{RP}^{o}=0.15\,V\]\[C{{r}^{3+}}+3e\xrightarrow{{}}Cr;E_{RP}^{o}=-0.74\,V\](Higher\[{{E}_{OP}}\]) \[\therefore \]\[{{E}_{cell}}=E_{P{{O}_{Cr}}}^{o}+E_{R{{P}_{Sn}}}^{o}=0.74+0.15=+0.89\,V\]You need to login to perform this action.
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