Given Fe2+ | Fe , Cr3+ | Cr, and Mn2+ | Mn half cell...?
Answer:
Each of your half-equations is written as a reduction, so the oxidising agent is on the left and the reducing agent on the right in each case. The best oxidising agent will be the species on the left of the equation with the most positive value for E°. The best reducing agent will be the species on the right of the equation with the most negative value for E°.
To work out the overall equation for the redox reaction you need to combine the half equations.
To use your example:
Fe2+ 2e- => Fe -0.44 V
Cr3+ 3e- => Cr -0.74 V
First look at the E° values - the one with the most positive (or least negative) value will go in the forward direction; the one with the least positive (or most negative) value will go in the reverse direction. So Fe2+ goes to Fe and Cr goes to Cr3+.
Now balance the electrons - multiply the first equation by 3 and the second by 2.
3Fe2+ 6e- => 3Fe -0.44 V
2Cr3+ 6e- => 2Cr -0.74 V
(note that the E° value does not change)
The add them together - the first one plus the reverse of the second one
3Fe2+ + 2Cr --> 3Fe + 2Cr3+
Ecell = -0.44 + 0.74 = +0.30V
(note that it is now +0.74 because we have reversed the second equation)
If you're doing problems like this I'm assuming you have a reduction potentials chart handy?
The half reactions are:
Fe2+ + 2e- --> Fe
Cr3+ + 3e- --> Cr
Mn2+ + 2+ --> Mn
The strongest oxidizer is the one that has the highest reduction potential (easily reduced)
The strongest reducer is the one that has the lowest reduction potential (easily oxidized)
When trying to figure out whether or not you have a reaction you compare the reduction potentials happening at each node.
Ered should be higher at the cathode (reduction) and the Ered should be lower at the anode (oxidation)
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