Question
What is the $$e.m.f$$ for the given cell ?
$$Cr\left| {C{r^{3 + }}\left( {1.0M} \right)} \right|\left| {C{o^{2 + }}\left( {1.0M} \right)} \right|Co$$
$$\left( {{E^ \circ }\,{\text{for}}\frac{{C{r^{3 + }}}}{{Cr}} = - 0.74\,volt} \right.$$ $$\left. {{\text{and}}\,\,{E^ \circ }\,{\text{for}}\frac{{C{o^{2 + }}}}{{Co}} = - 0.28\,volt} \right)$$
A.
$$ - 0.46\,volt$$
B.
$$ - 1.02\,volt$$
C.
$$ + 0.46\,volt$$
D.
$$1.66\,volt$$
Answer :
$$ + 0.46\,volt$$
Solution :
$$\eqalign{
& E_{\frac{{C{r^{3 + }}}}{{Cr}}}^ \circ = - 0.74\,V,E_{\frac{{C{o^{2 + }}}}{{Co}}}^ \circ = - 0.28\,V \cr
& {\text{The given cell reaction is}} \cr
& Cr\left| {C{r^{3 + }}\left( {1.0\,M} \right)} \right|\left| {C{o^{2 + }}\left( {1.0M} \right)} \right|Co \cr
& \therefore \,\,Cr\,\,{\text{is anode and }}Co{\text{ is cathode}} \cr
& E_{cell}^ \circ = E_C^ \circ - E_A^ \circ \cr
& = - 0.28 - \left( { - 0.74} \right) \cr
& = - 0.28 + 0.74 \cr
& = + 0.46\,V \cr} $$