Calculate the cell potential for a cell based on the reaction
Cu(s) + 2Ag+(aq) -> Cu2+(aq) + 2Ag(s)
when the concentrations are as follows:
[Ag+]= 0.53 M, [Cu2+] = 0.9 M.
(The temperature is 25◦C and E0 = 0.4624 V.)
Answer in V.
Calculate Ehalfcell for the REDUCTION half equation from
Ehalfcell = Eo-(0.05016/n)log [(red)/(ox)]
Then reverse the half cell that is oxidized, (reverse the sign, also), then add to the reduction half cell. That will be Ecell.
Post your work if you get stuck.
I got:
0.4624-(.05016/2) log(.80/.34)= 98.569
and i don't get the second part where you reverse it and add it...do i add it to 98.569??
To calculate the cell potential, we can use the Nernst equation:
E = E° - (0.0592 V / n) * log(Q)
Where:
- E is the cell potential
- E° is the standard cell potential (given as 0.4624 V)
- n is the number of electrons transferred in the balanced equation (in this case 2, since 2 electrons are transferred)
- Q is the reaction quotient, which is calculated using the concentrations of the species involved.
In this case, the reaction quotient (Q) can be calculated using the concentrations of Ag+ and Cu2+:
Q = [Cu2+] / [Ag+]^2
= 0.9 / (0.53)^2
Let's plug these values into the equation and solve for E:
E = 0.4624 V - (0.0592 V / 2) * log(0.9 / (0.53)^2)
Calculating this expression will give us the cell potential (E) in volts.