Need help with a question! I cut out irrelevant bits of it as it was a multi-part question so let me know if anything doesn't make sense/there's information missing :p
In acidic solution, ions containing titanium can react according to the half-equation TiO2+ (aq) + 2H+ (aq) + e- ⇌ Ti3+ (aq) + H2O (l) where E° = -0.06 V. In the diagram (attached), A and B are inert graphite electrodes, and all ions have a concentration of 1M. If both A and B electrodes are replaced with iron, what equation will now occur at Electrode B?
I wrote 2TiO2+ (aq) + 4H+ (aq) + Fe (s) -> 2Ti3+ (aq) + 2H2O(l) + Fe2+(aq) but this is wrong - the answer is instead supposed to be Fe(s) + 2H+(aq) -> Fe2+(aq) + H2(g).
I don't understand this - why is 2H+(aq) + 2e- -> H2(g) a half-equation that is considered? Is it because the half-cell containing electrode B has H+ ions in solution and the question states it is an acidic solution? But the half-equation contains H2 which isn't part of the question or in the galvanic cell so shouldn't it be eliminated as an option?
In a galvanic cell the strongest oxidant will always react with the strongest reductant.
In this case they gave you the E° value of TiO
2+ (aq) + 2H
+ (aq) + e
- ⇌ Ti
3+ (aq) + H
2O(l), which is -0.06 V. If you place this in the electrochemical series, you'll find that 2H
+(aq) is a stronger oxidant than TiO
2+. Because of this 2H
+(aq) will be reduced in preference (hence: 2H
+(aq) + 2e
- -> H
2(g)).
The reaction is considered because it's part of the half-cell containing electrode B. It doesn't matter that H
2 isn't part of the galvanic cell because its a product (if H
2 was a reactant than it would be a different matter). It only matters that H
2+ is being reduced. The H
2 is just going to be released into the air as a gas!