Please explain what you do and why.
Answer Question a: Potential of cell
First step: Identify the semi-reactions of oxidation and reduction and the balanced complete equation of cell
For the Co, to write the semi-reaction of oxidation you just write the inverse reaction, and change the sign of Eºred to obtain the Eºoxid value.
Second step: Determinate the Eº of the galvanic cell
*This is in standard conditions: 298 K and solutions 1M
Third step: Determinate the E of the galvanic cell if you use 250 mL of Al(NO3)3 0.15 M and 250 mL of Co(NO3)2 0.85 M
where
1 mol of Co(NO3)2 produces 1 mol of Co2+, then, [Co(NO3)2] = [Co2+] = 0.85 M
1 mol of Al(NO3)3 produces 1 mol of Al3+, then, [Al(NO3)3] = [Al3+] = 0.15 M
Answer Question b: mass gained/lost by cobalt electrode
First step: Calculate the equilibrium constant of the reaction:
then
Second step: Determinate the reaction direction
Third step: Determinate the amount of Co2+ produced
2 Al3+ | + | 3 Co | --> | 2 Al | + | 3 Co2+ | |
Initial | 0.15 | -- | -- | 0.85 | |||
Change | -2x | -- | -- | +3x | |||
Equilibrium | 0.15-2x | -- | -- | 0.85+3x |
the moles produced of Co2+ are (the cell has 250 mL = 0.25 L):
Fourth step: mass of Co lost
Answer Question c: current generated
where
then, the current generated is
RESUME
Please explain what you do and why. A galvanic cell is setup with the following two...
Draw the voltaic cell that will give the most positive cell potential choosing from the following half reactions: Eo (vs. SHE) Zn2+ (aq) + 2e- Zn (s), Eo= -0.76 Al3+ (aq) + 3e- Al (s), Eo= -1.66 Cr3+ (aq) + 3e- Cr (s), Eo= -0.74 Co2+ (aq) + 2e- Co (s), Eo= -0.28
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