Balance the following redox reaction in acidic solution using the 1/2 reaction m

Question

Balance the following redox reaction in acidic solution using the 1/2 reaction method. Show the two balanced half reactions with electrons before combining to form the whole reaction and the final overall balanced reaction. Fe2 +(aq) + MnO4(aq) rightarrow Fe3+(aq) + Mn2+(aq) (acidic solution) How long must a constant current of 50.0 A be passed through an electrolytic cell containing aqueous Cu2+ ions to produce 325g of copper metal? Give an answer in minutes. Time = min Determine the standard reduction potential (E) at 298K for the half reaction Cu_2S (s) + 2e- a 2Cu(s) + S degree (aq) E degree = Ksp for Cu_2 S is 2.51times 10 and E degree = +0.52V for Cu^-1 + e- Cu(s)

Explanation / Answer

Answer

B> Given Fe2+(aq) + MnO4-(aq) => Fe3+(aq) + Mn2+(aq) (acidic solution)
- Writing the half reactions:

Fe2+(aq) => Fe3+(aq) (oxidation)

MnO4-(aq) => Mn2+(aq) (reduction)

Balancing each half reaction:

Fe2+(aq) => Fe3+(aq) + e-

MnO4-(aq) + 8H+(aq) + 5e- => Mn2+(aq) + 4H2O

Combining the half reactions to give the overall reaction:

Fe2+(aq) + MnO4-(aq) + 8H+(aq) + 5e- => Fe3+(aq)+ Mn2+(aq) + 4H2O

overall balance equation

5Fe2+(aq) + MnO4-(aq) + 8H+(aq) => 5Fe3+(aq) + Mn2+(aq) + 4H2O

--D> given I= 50 A,

moles of copper metal produced = 325/63.5= 5.118 mol
Cu2+ + 2e-   >Cu

therefore to produce 1 mole of copper 2 mole of electrons are required.

Thus Q= 96500*2* 5.118 coloumb

Q=I*t
t= 96500*2*5.118 / (50*60)= 330 min

E> Given

Cu2S + 2e- ==> 2Cu(s) + S2-

Reactions are Cu + 1e- ==> Cu+
S+2e- ==> S2-

Using equation -- Ecell = E0 cell + (RT/nF) log K

using the values RT/F = 0.0591 V

Therefore E0cell = Ecell- (0.0591/2 ) log 2.51* 10-48

E cell= 0.52V-1.2 V+ 0.502

on solving E0 cell = + 0.898 V

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