For spontaneous Ni, Al galvanic cell, calculate Ecell if [Al3+]=1.80M, and [Ni2+]=0.0096M
For a spontaneous process, Ecell
and E0 both must be positive.
For spontaneous Ni, Al galvanic cell, calculate Ecell if [Al3+]=1.80M, and [Ni2+]=0.0096M
A voltaic cell consists of an Ni/Ni2+ half-cell and a Co/Co2+ half-cell. Calculate Ecell when conc. of Ni2+ = 0.142 M and of Co2+ = 0.468 M. The reduction potential for Ni2+ is -0.247 V and for Co2+ is -0.277 V. Answer: 0.01467 A voltaic cell consists of an Al/Al3+ half-cell and a Cd/Cd2+ half-cell. Calculate {Al3+} when {Cd2+} = 0.401 M and Ecell = 1.299 V.Use reduction potential values of Al3+ = -1.66 V and for Cd2+ = -0.40...
If you construct a Galvanic / Voltaic cell using a Ni(s) / Ni2+(aq) half cell and an Al(s)/ Al3+(aq)half cell,and connect the whole cell to a voltmeter by connecting the Al half cell to the positive terminal of the voltmeter, what will be the display?
Al3 (aq) + Ni(s). 3. Consider the unbalanced reaction Al(s)+ Ni2 (aq) > a. write balanced cathode half reaction b. write the balanced anode half reaction C. Write the balanced overall reaction d. Calculate E。, the cell potential. Is the reaction spontaneous as written? e Calculate Eolf [AP ] = 4.00 M, [Ni2+] = 2.00 x 10-5 M.
Question 4 A galvanic cell uses the reaction: 2Ag+(aq) + Ni(s) --> 2Ag(s) + Ni2+(aq) a) calculate the Standard Cell Potential (Ecell) of this galvanic cell and b) which metal is the cathode? a) 1.03 V; b) nickel a) 1.03 V; b) silver oooo a) 0.57 V; b) nickel a) 0.57V; b) silver
The value of Eocell for the cell shown below is +1.41 V. Al(s) | Al3+(aq) || Ni2+(aq) | Ni(s) The concentration of Al3+(aq) is 0.050 M and the concentration of Ni2+(aq) is 2.0 M. Estimate the value of Ecell at 25 oC. -1.41 V +1.34 V +1.38 V +1.48 V +1.44 V
Calculate the standard cell potential (∆Eo) for the
galvanic cell:
Ni (s) 1 Ni2+ (aq) II Ag+ (aq) 1 Ag (5) Given: E Half Reaction Ag+ (aq) +e- → Ag (s) Ni2+ (aq) + 2e- → Ni (s) 0.79 Volts -0.23 Volts
A voltaic cell consists of an Al/Al3+ half-cell and an Fe2+/Fe3+ half-cell. Calculate Ecell when conc. of Fe2+= 0.402 M, Fe3+= 0.073 M and Al3+= 0.239 M. Use the reduction potentials for Al3+ is -1.66 V and for Fe3+ is 0.77 V. How do you decide which is oxi, red.
A voltaic cell consists of an Al/Al3+ half-cell and a Cd/Cd2+ half-cell. Calculate {Cd2+} when {Al3+} = 0.306 M and Ecell = 1.27 V. Use reduction potential values of Al3+ = -1.66 V and for Cd2+ = -0.40 V.
Calculate the standard cell potential and determine if the reaction is spontaneous in the forward direction(as written). Identify each oxidizing agent and reducing agent. Ni (s) + Zn2+(aq) ----> Ni2+(aq) + Zn (s) Ni (s) + Pb2+(aq) -------> Ni2+(aq) + Pb (s) Al (s) + 3 Ag+(aq) ---------> Al3+(aq) + Ag (s) Pb (s) + Mn2+(aq) ----------> Pb2+(aq) + Mn (s)
15) Consider a galvanic cell with a Ni electrode in 1.0 M Ni2+(aq) joined by a salt bridge to a silver electrode in 1.0 M Ag+(aq). Calculate the concentrations of Ag+(aq) and Ni2+(aq) once the cell is "dead". Answer: [Ag+] = 4.6 * 10-18 M, [Ni2+] = 1.5 M