For the following cell, write the equations of the reactions occurring at the cathode and the anode, and an equation for the overall or net cell reaction. Calculate the standard cell potential. Cr(s) Cr2+ (aq). Sn2+ (aq) Sn(s)

2 months ago

Answered By Mandy H

For electrochemical cells, remember that the strongest reducing agent (SRA) is oxidized at the anode (SRAOA) and the strongest oxidizing agent (SOA) is reduced at the cathode (SOARC). Using your data booklet (pg. 7- table of selected standard electrode potentials) determine which is your SOA (start on the top left hand side, determine whether you see Cr, Cr^{2+}, Sn^{2+} or Sn first) and which is your SRA (start on the bottom right hand side, what do you see first).

You should have determined that the SOA is Sn^{2+ }and the SRA is Cr.

The reactions that happens:

cathode: Sn^{2+ }_{(aq)} + 2e^{-} --> Sn_{(s)}

anode: Cr_{(s)} --> Cr^{2+ }_{(aq) }+ 2e^{- }

To write your net, make sure that electrons 'cancel' on both sides of the equation (they do in this case). If they don't then you need to multiply by a coefficient.

To calculate standard cell potential there are two formulas, but both will give you the exact same answer.

Reduction happens at the cathode (tin(ll) E_{reduction=} =0.14) and oxidation happens at the anode (chromium E_{reduction} = -0.91)

Before using this first equation, change the reduction into oxidation by switching the sign (remember the value given is for a reduction half reaction).

E_{net} = E_{reduction }- E_{oxidation}

= 0.14 - (+0.91)

= -0.77

If using this second equation, do NOT switch any of the signs as you are using the reduction potential already.

2 months ago

## Answered By Mandy H

For electrochemical cells, remember that the strongest reducing agent (SRA) is oxidized at the anode (SRAOA) and the strongest oxidizing agent (SOA) is reduced at the cathode (SOARC). Using your data booklet (pg. 7- table of selected standard electrode potentials) determine which is your SOA (start on the top left hand side, determine whether you see Cr, Cr

^{2+}, Sn^{2+}or Sn first) and which is your SRA (start on the bottom right hand side, what do you see first).You should have determined that the SOA is Sn

^{2+ }and the SRA is Cr.The reactions that happens:

cathode: Sn

^{2+ }_{(aq)}+ 2e^{-}--> Sn_{(s)}anode: Cr

_{(s)}--> Cr^{2+ }_{(aq) }+ 2e^{- }To write your net, make sure that electrons 'cancel' on both sides of the equation (they do in this case). If they don't then you need to multiply by a coefficient.

Net cell reaction: Sn

^{2+ }_{(aq)}+ Cr_{(s)}--> Sn_{(s) }+ Cr^{2+ }_{(aq)}To calculate standard cell potential there are two formulas, but both will give you the exact same answer.

Reduction happens at the cathode (tin(ll) E

_{reduction=}=0.14) and oxidation happens at the anode (chromium E_{reduction}= -0.91)Before using this first equation, change the reduction into oxidation by switching the sign (remember the value given is for a reduction half reaction).

E

_{net}= E_{reduction }- E_{oxidation}= 0.14 - (+0.91)

= -0.77

If using this second equation, do NOT switch any of the signs as you are using the reduction potential already.

E

_{net }= E_{red. cathode}+ E_{red anode}= (0.14) + (-0.91)

= -0.77