Which of the following reactions will occur spontaneously as written?
2. 3Sn{4+} (aq) + 2Cr(s) = 2Cr{3+} (aq) + 3Sn{2+}(aq)
3. 3Fe(s) + 2Cr{3+}(aq) = 2Cr(s) + 3Fe{2+}(aq)
4. Sn{4+}(aq) + Fe{2+}(aq) = Sn{2+}(aq) + Fe(s)
5. Sn{4+}(aq) + Fe{3+}(aq) = Sn{2+}(aq) + Fe{2+}(aq)
Can someone show me how to write them out to see if they will occur spontaneously thanks
Chem homework is to stressful
Answer:
1)Fe2+(aq)+2e-=Fe(s) redox potential -0.44 V
.2Fe2+(aq)=2Fe3+(aq)+2e- redox potential 0.77V but need to mult by 2 because 2 moles reacted
thus... 1.54V-0.44V=1.10 V spontaneous
2) 3Sn4+(aq)+6e-=3Sn2+ redox potential 0.15 V (need to mult by 3)
.2Cr=2Cr3+(aq)+6e- redox potential 0.74V mult by
1.48V+0.45V=1.93V spontaneous
3)3Fe=3Fe2+(aq)+6e- redox potential 0.44V (sign flip because reaction went the other way from number one)(mult by 3)
...2Cr3+(aq)+6e-=2Cr redox potential -0.74 V (see above half reaction for sign flip reasoning)(mult by 2)
1.32V-1.48V=-0.16 V not spontaneous
4)Sn4+(aq)+2e-=Sn2+ redox potential 0.15V
.Fe2+(aq)+2e-=Fe redox potential -0.44V
0.15V-0.44V=-0.29V not spontaneous
5) not balanced properly
positive redoex potential for reaction is spontaneous, negative is not
Divide each into a reduction and an oxidation.
Look up the electrochemical reduction potentials corresponding to each. Reverse the sign of the oxidation and add them together. Then decide the spontaneity based on the sign of the overall potential.
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