Find the emf of the cell described by the cell diagram Fe | Fe^2+ (1.500M) || Au

Question

Find the emf of the cell described by the cell diagram Fe | Fe^2+ (1.500M) || Au^3+ (0.00400M) | Au. A) 1.99 V B) 1.89 V C) 1.94 V D) 1.66 V E) 1.91 V Which one of the following reactions must be carried out in an electrolytic cell rather than in a galvanic cell? A) Zn^2+ (aq) + Ca(s) rightarrow Zn(s) + Ca^2+ (aq) B) Al^3+ (aq) + 3Br^- (aq) rightarrow Al(s) + (3/2)Br_2 (1) C) 2Al(s) + 3Fe^2+ (aq) rightarrow 2Al^3+ (aq) + 3Fe(s) H_2(g) + I_2(s) rightarrow 2H^+ (aq) + 2I^-(aq) E) Fe^2+ (aq) + Mg(s) rightarrow Fe(s) + Mg^2+ (aq) Select True or False: Under acidic conditions the correctly balanced redox reaction for MnO_4^-(aq) + C_2O_4^2-(aq) rightarrow Mn^2+(aq) + CO_2(aq) is: 2MnO_4^-(aq) + 5C_2O_4^2-(aq) + 8H^+ (aq) rightarrow 2Mn^2+ (aq) + 10CO_2(aq) + 8H_2O(1) A) True B) False

Explanation / Answer

20)

Lets find Eo 1st

from data table:

Eo(Fe2+/Fe(s)) = -0.44 V

Eo(Au3+/Au(s)) = 1.52 V

here:

cathode is (Au3+/Au(s))

anode is (Fe2+/Fe(s))

The chemical reaction taking place is

2Au3+(aq) + 3Fe(s) --> 2Au(s) + 3Fe2+(aq)

Eocell = Eocathode - Eoanode

= (1.52) - (-0.44)

= 1.96 V

Number of electron being transferred in balanced reaction is 6

So, n = 6

Use:

E = Eo - (2.303*RT/nF) log {[Fe2+]^3/[Au3+]^2}

2.303*R*T/n = 2.303*8.314*298.0/F= 0.0591

E = Eo - (0.0591/n) log {[Fe2+]^3/[Au3+]^2}

E = 1.96 - (0.0591/6) log (1.5^3/0.004^2)

E = 1.96-(0.052)

E = 1.91 V

Answer: E

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