Redox Reactions And Electrochemistry Ques 570
Question: Use the following standand electrode potentials, calculate $ \Delta G{}^\circ $ in kJ/ mol for the indicated reaction:
$ 5C{e^{4+}}(aq)+M{n^{2+}}(aq)+4H_2O(l)\xrightarrow{{}} $ $ 5C{e^{3+}}(aq)+MnO_4^{-}(aq)+8{H^{+}}(aq) $ $ MnO_4^{-}( aq )+8{H^{+}}( aq )+5{e^{-}}\xrightarrow{{}} $ $ M{n^{2+}}(aq)+4H_2O(l);E{}^\circ =+1.51V $ $ C{e^{4+}}(aq)+{e^{-}}\xrightarrow{{}}C{e^{3+}}(aq);E{}^\circ =+1.61,V $
Options:
A) $ -9.65 $
B) $ -24.3 $
C) $ -48.25 $
D) $ -35.2 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ E_{cell}^{{}^\circ }=E_RP^{{}^\circ }( RHS )-E_RP^{{}^\circ }( LHS ) $ $ =1.61-1.51\Rightarrow 0.10V $ $ \Delta G{}^\circ =-nFE{}^\circ \Rightarrow -5\times 96500\times 0.10J $ $ \Delta G{}^\circ =-48.25kJmo{l^{-1}} $
