Redox Reactions And Electrochemistry Ques 557
Question: Standard electrode potential for $ S{n^{4+}}/S{n^{2+}} $ couple is + 0.15 V and that for the $ C{r^{3+}}/Cr $ couple is $ -0.74V $ . These two couples in their standard state are connected to make a cell. The cell potential will be:
Options:
A) +1.19 V
B) +0.89 V
C) +0.18 V
D) +1.83 V
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Given $ {E_{S{n^{4+}}/S{n^{2+}}}}=+0.15V $
$ {E _{C{r^{3+}}/Cr}}=-0.74V $
$ E _{cell}^{{}^\circ }=E _{cathode}^{{}^\circ }-E _{anode}^{{}^\circ } $ $ =0.15-( -0.74 )=+0.89V $
