1-ele2 Calculate the cell potential for this reaction, in the direction shown, at 298 K. Zn²+ (0.0000270 M) → Zn²- 1.01 × 10° M E : V In this zinc concentration cell, the reaction would proceed spontaneously in th v direction shown reverse direction

Subject: chemistry

If the concentration of Ag+ is 0.0115 M, the concentration of H+ is 0.355 M, and the pressure of H2 is 1.00 atm, calculate the cell potential at 25.0°C.

Calculate the cell potential for this reaction, in the direction shown, at 298 K. 2+ Hg²+ (0.0000490 M) → Hg²+ 1.37 × 10⁰. E In this mercury concentration cell, the reaction would proceed spontaneously in the [ V

Consider the electrolysis reactions cathode anode whose standard reduction potentials can be found in this table. Calculate the voltage needed to drive the net reaction if the current is negligible. Ecell = Cu²+ (aq, 0.177 M) +2e=Cu(s) H₂O(1) ⇒ 10₂(g, 1 bar) + 2 H+ (aq, 0.195 M) + 2 e¯ The cell has a resistance of 66.0 and a current of 7.17 mA when the cathode overpotential is 0.231 V and the anode overpotential is 0.050 V. Calculate the voltage needed to overcome these effects and drive the net reaction. Ecell = V V

What is the calculated value of the cell potential at 298 K for an 2+ Cu²+ concentration is 3.83 × 10-4 M and the A1³+ concentration is 1.12 M? Ecell electrochemical cell with the following reaction, when the 2+ 3Cu²+ (aq) + 2Al(s) → 3Cu(s) + 2A1³+ (aq) V The cell reaction as written above spontaneous nonspontaneous for the concentrations given.

The free energy change for the following reaction at 25 °C, when [Pb²+] = 3.97×10-³ M and [Cu²+] = 1.20 M, is 104 kJ: Pb²+(3.97x10-³ M) + Cu(s)- → Pb(s) + Cu²+(1.20 M) AG = 104 kJ What is the cell potential for the reaction as written under these conditions? Answer: V Would this reaction be spontaneous in the forward or the reverse direction?

The standard cell potential for the reaction shown here is .653 V. Calculate the cell potential at 25°C with [Hg^2+] = 0.0500 M, [V^2+] = 0.100 M, [Hg2^2+] = 0.00500 M and [V^3+] = %3D 0.0200 M. [Hg?+] do.o500 M [V2+] = 0.100 M H0.0500 M [Hg* = 0.00500 M [V³*] = 0.0200 M 2 Hg²+ (aq) + 2 V²+ (aq) = Hg2²+ (aq) + 2 V³+ (aq)