Calculate The Cell Potential For The Reaction As Written At 25 00 C Given That Cr2 0 836 M And Ni2 1 0 0160 M 1 (350.57 KiB) Viewed 29 times
Cant figure out why I am wrong, please help? I am doing:
.50 - (8.314K*298K/2*96485)(ln 0.836/.0160) and the 96485 is the constant F Achieve provides to me.
Thank you!
Calculate the cell potential for the reaction as written at 25.00 °C, given that [Cr2+] = 0.836 M and [Ni2+1 = 0.0160 M. Use the standard reduction potentials in this table. S Cr(s) + Ni2+ (aq) + Cr2+ (aq) + Ni(s) E= | $.449 I I PC TOOLS x109
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