A redox reaction has an E degree_cell = 1.21 V. What can you conclude about the

Question

A redox reaction has an E degree_cell = 1.21 V. What can you conclude about the equilibrium constant (K) for the reaction? K 1 Nothing can be concluded about K from E degree_cell. Use appropriate data to determine which statement is true of the voltaic cell picture here. Cr is the anode; Fe is the cathode; electrons flow from right to left Cr is the cathode; Fe is the anode; electrons flow from right to left Cr is the cathode; Fe is the anode; electrons flow from left to right Cr is the anode; Fe is the cathode; electrons flow from left to right

Explanation / Answer

We know that at equilibrium condition the Ecell will be equal to zero

So the Nernst equation

E0cell = Ecell - 0.0592 / n log Kc

will become

E0cell = [-0.0592 / n] [log Kc]

Given E0cell = 1.21

1.21 X n / 0.0592 = logKc

20.43 n = log Kc

n> or = 1 (in all cases)

so Kc > 10^20

So Kc > 1

b) The nature of electrode to behave as cathode or anode depends upon the standard reduction potential values

If the standard reduction potential value of one electrode is lesser then it will act as anode and will undergo oxidation. It will give electrons to other electrode

Here the E0 of Cr is lesser than that of Fe so Chromium will act as andoe

The electrode metal with higher reduction potential will act as cathode and will undergo reduction. It will accept electron from other electrode

So Iron will act as cathode

The true statement will be

Cr will act as anode, Fe is the cathode and electrons will flow from right to left [Cr to Fe]

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