It appears that both of these recations are reduced, but the solution states tha

Question

It appears that both of these recations are reduced, but the solution states that the first equation is actually an oxiditation reaction (the anode part) and this was demonstrated by flipping the chemical reaction. How will I be able to determine which is the cathode and anode between two reactions that appear this way?

Consider the following half-reactions and their standard reduction potentials then give the standard line (cel notation for a voltaic cell built on these half reactions. Mn2+ aq) 2 e Mn(s) E 1.18 V 3+ Fe aq) 3 e Fe(s) E 0.036 V

Explanation / Answer

It is convention to write the haf cell reactions as reduction and the potentials as SRP values only.

However when you were given two half cells , just like the example above, see the potential values.

Mn has -1.18V abd Fehas -.-0036V

The negative sig n indicates both are reluctant to undergo reduction.

see the values .

Mn has more negative value which means it strongly refused to undergo reduction compared to Fe.

So Mn under goes oxidation and Fe undergoes reduction.

As a general rule if you have one half cell with positive SRP and other with negative SRP then the half cell with positive SRP undergoes reduction. (positive means readily undergo reduction)

If both are positive SRP, the one with higher positive value under goes reduction, other oxidation.

If both are negative SRP , the one with less negative value under goes reduction.

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