Question

Part 1.)

(a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in strong acid: Bro3 , H2O2, Mn2+, Ce4+, H2SO3 oBro3" он20, O Mn2+ o Ce4+ O H2SO3 OBro3" OBro3 OBro3 ОН20, он20, он20, < O Mn2+ < oMn2+ < OMn2+ o Ce4+ о се4+ oce4+ H2SO3 H2SO3 H2SO3 OBro3 он202 < омn2+ o Ce4+ H2SO3 (b) Assuming standard conditions, arrange the following in order of increasing strength as reducing agents in strong base: Sn2+, H2, Sn, F, Na o sn2t o sn2t он, он, o sn so sn o sn2t o sn2t он, он, < osn so sn < ООООО

Part 2.)

A voltaic cell operating under standard conditions (1.0 M ion concentrations) utilizes the following reaction:

2 Al(s) + 3 Ni²⁺(aq) 2 Al³⁺(aq) + 3 Ni(s)

What is the effect on the cell emf of each of the following changes?

(a) Water is added to the cathode compartment, diluting the solution:
The positive cell emf rises, becoming more positive. The positive cell emf drops closer to zero. The negative cell emf rises closer to zero. The negative cell emf drops, becoming more negative. No change in cell emf occurs.

(b) The size of the nickel electrode is increased:
The positive cell emf rises, becoming more positive. The positive cell emf drops closer to zero. The negative cell emf rises closer to zero. The negative cell emf drops, becoming more negative. No change in cell emf occurs.

(c) A solution of 1.0 M Al(NO₃)₃ is added to the anode compartment:
The positive cell emf rises, becoming more positive. The positive cell emf drops closer to zero. The negative cell emf rises closer to zero. The negative cell emf drops, becoming more negative. No change in cell emf occurs.

(d) Some Na₂S is added to the Al³⁺ compartment, precipitating some Al³⁺ as aluminum sulfide:
The positive cell emf rises, becoming more positive. The positive cell emf drops closer to zero. The negative cell emf rises closer to zero. The negative cell emf drops, becoming more negative. No change in cell emf occurs.

Answer 1

The order can be obtained from standard potential tables, and, for the oxidants it is:

$Mn^{2&plus;}< H_{2}SO_{3}< BrO_{3}^{-}< Ce^{4&plus;}< H_{2}O_{2}$

And for reductants:

$F^{-}< Sn^{2&plus;}< H_{2}< Sn< Na$

The equation that represents the emf of he cell is the Nernst equation:

$E=E^{o}-\frac{RT}{nF}\cdot ln\left ( \frac{[Al^{3&plus;}]^{2}}{[Ni^{2&plus;}]^{3}} \right )$

If concentrations are both 1 M, the ln(1) = 0, and the emf is the same as the standard potential for the cell, which is:

$E^{o}=E^{o}_{red}-E^{o}_{ox}=-0.25V-(1.66V)=1.41V$

a) If water is added to the cathode (where the reduction happens), the concentration of Ni²⁺ would decrease, which would make the ln increase, so the overall E would decrease due to the negative sign of the second term of the equation, so the positive emf comes closer to zero.

b) If the amount of solid Ni changes, there are no changes in the emf, since the solid reactants do not participate in the equilibrium reaction.

c) If Al³⁺ is added to the anode in the same concentration as the original, there is no overall change in the concentration of Al³⁺, which means that there are no changes in the potential of the cell.

d) If Al³⁺ precipitates, its concentration decreases, which decreases the value of the ln, consequently augmenting the total potential of the cell, which becomes more positive.

Mn2+ <H2SO3 < Bro< Cett <H2O

F- <Sn2+ <H <Sn < Na

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E" = E - E = -0.25V - (1.66V) = 1.41V