Redox Reactions And Electrochemistry Ques 601
Question: What is the e.m.f for the given cell-
$ Cr|C{r^{3+}}(1.0M)||C{o^{2+}}(1.0M)|Co $ $ (E{}^\circ forC{r^{3+}}/Cr=-0.74 $ volt and $ E{}^\circ $ for $ C{o^{2+}}/Co=-0.28 $ volt)
Options:
A) $ -0.46 $ volt
B) $ -1.02,~volt $
C) $ +0.46 $ volt
D) 1.66 volt
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ E_{C{r^{3+}}/Cr}^{{}^\circ }=-0.74V,,E_{C{o^{2+}}/Co}^{{}^\circ }=-0.28V $ The given cell reaction is $ Cr|C{r^{3+}}(1.0M)||C{o^{2+}}(1.0M)|Co $
$ \therefore Cr $ is anode and Co is cathode $ E_{cell}^{{}^\circ }=E_C^{{}^\circ }-E_A^{{}^\circ }=-0.28-(-0.74) $ $ =-0.28+0.74=+0.46V $
