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 $



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