One half-cell in a voltaic cell is constructed from a silver wire electrode in a .25 M solution of AgNO3 . The other half-cell consists of a zinc electrode in a .001 M solution of Zn(NO3)2. Calculate the cell potential
I have no idea and am so stressed, Help please
Look up reduction potentials for these two.
E = Eo - (0.05916/n)*log[(red)/(ox)] for Ag.
E = Eo - etc etc for Zn.
Reverse one of the Eo values(change sign, too) and add to the other. Do this in such a fashion that the Ecell = + value volts. That gives you the Eocell.
Don't worry, I'm here to help! To calculate the cell potential of a voltaic cell, we can use the Nernst equation. This equation relates the cell potential to the concentrations of the ions in the half-cells.
The Nernst equation is given by:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
- Ecell is the cell potential
- E°cell is the standard cell potential, which can be obtained from a table of standard reduction potentials
- n is the number of electrons transferred in the balanced equation of the cell reaction
- Q is the reaction quotient, which is given by the concentrations of the ions in the half-cells
First, we need to write the balanced equation for the cell reaction. The half-reactions are:
Ag+ + e- -> Ag
Zn -> Zn2+ + 2e-
By inspecting these half-reactions, we can see that 1 electron is transferred.
Now, we need to calculate the reaction quotient (Q) using the concentrations of the ions.
Q = [Ag+]/[Zn2+]
= (0.25 M) / (0.001 M^2)
Now, we can look up the standard reduction potentials for the half-reactions:
E°cell = E°(cathode) - E°(anode)
From a table, we find that:
E°(cathode) = 0.80 V (for the reduction of Ag+ to Ag)
E°(anode) = -0.76 V (for the oxidation of Zn to Zn2+)
Substituting the values into the Nernst equation, we get:
Ecell = 0.80 V - (0.0592/1) * log(0.25 M / (0.001 M)^2)
Now, you can plug in the values into the equation and calculate the cell potential.