Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: sharonjh on September 11, 2010, 06:59:44 AM
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Consider an electrochemical cell with the following cell reaction where all reactants and products are at standard-state conditions.
Cu2+ (aq) + H2(g) ® Cu(s) + 2H+(aq)
Predict the effect on the emf of this cell of adding NaOH solution to the hydrogen half-cell until the pH equals 7.0.
The emf will increase.
The emf will decrease.
No change in the emf will be observed.
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Have you come up with any ideas?
Try to write the expression for the potential of the hydrogen half-cell...
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The potential for H2 is 0 so does this mean the NaOH will have no affect on the EMF?
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The potential for H2 is 0 so does this mean the NaOH will have no affect on the EMF?
There's no point in guessing the answer...
Try to write the expression for the potential of the hydrogen half-cell.
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The potential for H2 is 0 so does this mean the NaOH will have no affect on the EMF?
There's no point in guessing the answer...
Try to write the expression for the potential of the hydrogen half-cell.
2H+(aq) + 2e- → H2(g) v=0.00
Do I need to use the Nernst equation?
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2H+(aq) + 2e- → H2(g) v=0.00
Do I need to use the Nernst equation?
Yes, that's exactly what you have to use... try to find out how E changes when [H+ (to be precise you should have written H3O+) decreases (due to the NaOH added).
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2H+(aq) + 2e- → H2(g) v=0.00
Do I need to use the Nernst equation?
Yes, that's exactly what you have to use... try to find out how E changes when [H+ (to be precise you should have written H3O+) decreases (due to the NaOH added).
Great...thank you so much for your *delete me*