What is the overall cell potential for this redox reaction?
ba + ca2+ _ ba2+ + ca
given: ba2+ + 2e_ _ ba(s) _2.90 and ca2+ + 2e_ _ ca(s) _2.87

I hope this problem I will provide might help you solve your problem:

What is the overall cell potential for this redox reaction?

Ca2+ + 2Li = Ca + 2Li+ 
Ca2+ + 2e- = Ca -2.87 V 
Li = Li+ + e- -3.05 V 

ANSWER:

 Ecell = Eox + Ered 

The oxidation half-cell reaction is Li = Li+ + e-  
and the given voltage is IN FACT the voltage for the reduction reaction of lithium. Since we have an oxidation reaction, we must reverse the sign of the given voltage to get the voltage of the oxidation reaction. Hence, Eox = -Ered = -( -3.05 V) = + 3.05 V 

The reduction half-cell reaction is Ca2+ + 2e- = Ca Ered = -2.87 V 

So, Ecell = 3.05 V + (-2.87 V) = 0.18 V


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