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Size:  161.5 KBCan someone help me with this question? I'm having my A2 this friday and would appreciate any help given. For part (i) why is the Ecell calculated as 0.40-(-0.83)= +1.23V? Based on the data booklet, 2H2O +2e ----> H2 + 2OH : E= -0.83V. isn't the electrode reaction for cathode change sign to +0.83V since it's reversed?
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    No, you keep the standard potential as the sign it's given, it will become positive in the equation lhs-rhs = E
 
 
 
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