Cr(s) ==> Cr3+ + 3e Eo = 0.74v = E1
Ce^4+ + e ==> Ce^3+ Eo = 1.61v = E2
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Cr(s) + 3Ce^4+ ==> 3Ce^3+ + Cr^3+
Eocell = E1+E2 = ? for standard conditions; i.e., 1 M concn at 25C.
For the second part, you want to use this.
Ecell = EoCell - (0.0592/n)log Q and
log Q = ((Ce^3+)(Cr^3+)/(Ce^4+)(Cr)(s)
Substitute the concns given in the problem and solve for Ecell.
A voltaic cell utilizes the following reaction and operates at 298K
3Ce{4+}(aq)+Cr(s)--->3Ce{3+}(aq)+Cr{3+}(aq)
{}=Charge
What is the EMF of this cell under standard conditions?
What is emf of cell when [Ce{4+}]=2.1M, [Ce{3+}]=.13M, and [Cr{3+}]=2.0*10^-2
1 answer