1) You need an acid that will spontaneously oxidize Sn, but NOT spontaneously ox

Question

1) You need an acid that will spontaneously oxidize Sn, but NOT spontaneously oxidize Ag. How many of the following acids meet these criteria?:
Hydrochloride acid, iodic acid, nitric acid, periodic acid
a) 1 b) 2 c) 0 d) 3 e) 4

2) What is deltaG for the balanced disproportionate reaction of IO3- under standard conditions?
a) 386 kJ b) 77.2 kJ c) 193 kJ d) -2.70 x 10^3 kJ e) -193 kJ
3) What is the deltaG for the spontaneous reaction that will occur when Cr wire is dipped into an aqueous solution of CoCl2 under standard condition?
a) -88.8 kJ b) -133 kJ c)-266 kJ d) -405 kJ e) -608 kJ f)-1.22 x 10^3 kJ 1) You need an acid that will spontaneously oxidize Sn, but NOT spontaneously oxidize Ag. How many of the following acids meet these criteria?:
Hydrochloride acid, iodic acid, nitric acid, periodic acid
a) 1 b) 2 c) 0 d) 3 e) 4

2) What is deltaG for the balanced disproportionate reaction of IO3- under standard conditions?
a) 386 kJ b) 77.2 kJ c) 193 kJ d) -2.70 x 10^3 kJ e) -193 kJ
3) What is the deltaG for the spontaneous reaction that will occur when Cr wire is dipped into an aqueous solution of CoCl2 under standard condition?
a) -88.8 kJ b) -133 kJ c)-266 kJ d) -405 kJ e) -608 kJ f)-1.22 x 10^3 kJ
Hydrochloride acid, iodic acid, nitric acid, periodic acid
a) 1 b) 2 c) 0 d) 3 e) 4

2) What is deltaG for the balanced disproportionate reaction of IO3- under standard conditions?
a) 386 kJ b) 77.2 kJ c) 193 kJ d) -2.70 x 10^3 kJ e) -193 kJ
3) What is the deltaG for the spontaneous reaction that will occur when Cr wire is dipped into an aqueous solution of CoCl2 under standard condition?
a) -88.8 kJ b) -133 kJ c)-266 kJ d) -405 kJ e) -608 kJ f)-1.22 x 10^3 kJ

Explanation / Answer

3) Half reactions:

Cr3+ + 3e- ---> Cr. Eº = -0.74V; Co2+ + 2e- ---> Co. Eº= -0.28V

Your first reaction has a DGº1 and the second has a DGº2. The overall reaction needs both half reactions, so that

2Cr + 3CoCl2 --> 2CrCl3 + 3Co, with DGº3.

Then, DGº3 = 3DGº2 - 2DGº1... Consider that DG= -nFEº, in both cases

DGº3 = 3(2)(96485C)(-0.28V) - (-2)(3)(96485C)(-0.74V) = -266298.6 J.

The answer then is -266 kJ

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