Redox Reactions And Electrochemistry Ques 710
Question: Emf of the cell
$ Ni|N{i^{2+}}(0.1,M)||A{u^{3+}},(1.0,M)|,Au $ will be $ (E_{Ni/N{i^{2+}}}^{{}^\circ }=0.25,,E_{Au/A{u^{3+}}}^{{}^\circ }=1.5,V) $
Options:
A) 1.75 V
B) $ + $ 1.7795 V
C) $ + $ 0.7795 V
D) $ - $ 1.7795 V
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Cell reaction: $ 3Ni+2A{u^{+3}}\xrightarrow{{}}3N{i^{+2}}+2Au $
$ E_{cell}=E_{cell}^{{}^\circ }-\frac{0.0591}{6}\log ,\frac{{{[N{i^{+2}}]}^{3}}}{{{[A{u^{+3}}]}^{2}}} $
$ =(0.25+1.5)-\frac{0.0591}{6}\log \frac{{{(0.1)}^{3}}}{{{(1)}^{2}}} $
$ =1.75-\frac{0.0591}{2}\log ,(.1),=,1.75+0.295=+1.7795,V $
