For the cell shown, the measured cell potential, Eceil, is -0.3675 V at 25 °C. Pt(s) | H, (8,0.889 atm) | H* (aq, ? M) |

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For The Cell Shown The Measured Cell Potential Eceil Is 0 3675 V At 25 C Pt S H 8 0 889 Atm H Aq M 1 (23.68 KiB) Viewed 64 times

For The Cell Shown The Measured Cell Potential Eceil Is 0 3675 V At 25 C Pt S H 8 0 889 Atm H Aq M 2 (19.15 KiB) Viewed 64 times

For The Cell Shown The Measured Cell Potential Eceil Is 0 3675 V At 25 C Pt S H 8 0 889 Atm H Aq M 3 (28.67 KiB) Viewed 64 times

For The Cell Shown The Measured Cell Potential Eceil Is 0 3675 V At 25 C Pt S H 8 0 889 Atm H Aq M 4 (16.06 KiB) Viewed 64 times

For the cell shown, the measured cell potential, Eceil, is -0.3675 V at 25 °C. Pt(s) | H, (8,0.889 atm) | H* (aq, ? M) || Cd2+(aq, 1.00 M) | Cd(s) The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, Eº, are 2 H+ (aq) + 24H,() E° = 0.00 V Cd + (aq) + 2 Ca(s) E° = -0.403 V Calculate the H concentration.
Calculate the cell potential for the reaction as written at 25.00 °C, given that (Zn²+] = 0.864 M and (Sn²+] = 0.0100 M. Use the standard reduction potentials in this table. Zn(s) + Sn?+(aq) = Zn”+ (aq) +Sn(s) E V
The amount of (aq) in a solution can be determined by titration with a solution containing a known concentration of S, 0- (aq) (thiosulfate ion). The determination is based on the net ionic equation 25,0?" (aq) + 13(aq) 5,0"(aq) + 31" (aq) Given that it requires 27.0 mL of 0.500 M Na,8,0, (aq) to titrate a 15.0 mL sample of T; (aq), calculate the molarity of TƯ (aq) in the solution. 115) = M
The voltage generated by the zinc concentration cell described by the line notation Zn(s) | Zn2+(aq,0.100 M) || Zn2+(aq, ? M) | Zn(s) is 14.0 mV at 25 °C. Calculate the concentration of the Zn2+ (aq) ion at the cathode.