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All Message Boards : Cell potential
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 Message 1 of 2 in Discussion 
From: MSN NicknameAlbert1145  (Original Message)Sent: 10/22/2008 9:32 AM
Hi Steve,

Q1) The galvanic cell described by:
Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)
has a cell potential of 1.101 volts. Given that Zn(s) -> Zn2+(aq) + 2e- has an oxidation potential of 0.762 volts, determine the reduction potential for Cu2+ in volts.

Is it just 1.101 - 0.762 = 0.388??


Q2)
Given this system:
3Ag(s) + NO3-(aq) + 4H+(aq) -> 3Ag+(aq) + NO(g) + 2H2O(l)
Anode reaction:
Ag(s) -> Ag+(aq) + 1e- Eo = -0.799 V
Cathode reaction:
NO3-(aq) + 4H+ + 3e- -> NO(g) + 2H2O(l) Eo = 0.964 V
Determine the standard cell potential in volts.

Is it 0.964 + 0.799 = 1.763??

They look too easy so I don't think I got them right. Are they?

Thanx


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Reply
 Message 2 of 2 in Discussion 
From: MSN Nickname·Steve·Sent: 10/22/2008 10:07 PM
1)  Hi Albert, that's right, 1.101 V �?0.762 V  =  +0.339 V.
 
2)  E°cell  =  �?/FONT>0.799 V  +  0.964 V  =  +0.165 V (don't forget the negative sign)
 
Just remember,  E°cell  =  E°ox  +  E°red
and, the oxidation half-reaction occurs at the anode (�?, and the reduction half-reaction occurs at the cathode (+).