Calculate the equilibrium constant for each of the reactions at 25 ∘C.
2Fe3+(aq)+3Sn(s)→2Fe(s)+3Sn2+(aq)
O2(g)+2H2O(l)+2Cu(s)→4OH−(aq)+2Cu2+(aq)
Br2(l)+2I−(aq)→2Br−(aq)+I2(s)

Question

Calculate the equilibrium constant for each of the reactions at 25 ∘C.
2Fe3+(aq)+3Sn(s)→2Fe(s)+3Sn2+(aq)
O2(g)+2H2O(l)+2Cu(s)→4OH−(aq)+2Cu2+(aq)
Br2(l)+2I−(aq)→2Br−(aq)+I2(s)

anonymous · Answer

What information are you given about the concentrations of the reactants and products? With what you have shown you can write the equilibrium expressions, but you need more information to actually calculate the values of the equilibrium constants.

anonymous · Answer

there is no information given. the topic is electrochemistry.

anonymous · Answer

You can not calculate the values of the equilibrium constants if you do not know (or can not determine) the equilibrium concentrations.

joannablackwelder · Answer

You can, using the cell potential values.$$K= 10^{nE ^{o}/.05916}$$ where is is the number of moles of electrons in the balanced half reactions.

joannablackwelder · Answer

*where n is

joannablackwelder · Answer

http://www.sfu.ca/chemistry/groups/Li/chem215/ECmeasure.PDF

jfraser · Answer

if you know the reduction potentials, you can calculate the equilibrium constant, as @joannablackwelder shows