For Part A, we’re being asked if the given redox reaction occurs spontaneously in the forward direction.
Recall that the greater the E°cell of a redox reaction, the more likely the reaction will occur (more spontaneous reaction)
E°cell > 0 → spontaneous in the forward direction
E°cell < 0 → nonspontaneous in the forward direction
We will have to calculate for the cell potential to determine the spontaneity of the redox reaction using the following steps:
Step 1. Write the two half-cell reactions
Step 2. Identify the oxidation half-reaction (anode) and the reduction half-reaction (cathode)
Step 3. Calculate E°cell.
Recall that standard reduction potential for the Standard Hydrogen Electrode:
2 H+(aq) + 2 e- → H2(g) E° = 0.000 V
Let's take a look at each reaction given:
Part A. Predict whether the following reactions are spontaneous
Pd2+(aq) + H2(g) → Pd(s) + 2 H+(aq) Pd2+(aq) + 2 e- → Pd(s) E° = 0.987 V
Sn4+(aq) + H2(g) → Sn2+(aq) + 2 H+(aq) ) Sn4+(aq) + 2 e- → Sn2+(aq) E° = 0.154 V
Ni2+(aq) + H2(g) → Ni(s) + 2 H+(aq) Ni2+(aq) + 2 e- → Ni(s) E° = -0.25 V
Cd2+(aq) + H2(g) → Cd(s) + 2 H+(aq) Cd2+(aq) + 2 e- → Cd(s) E° = -0.403 V
From your answers, decide which of the above metals could be reduced by hydrogen
Part B. Identify the oxidizing agents and reducing agents in the reaction in Part A
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