The emf of the following cell at 25° is 0.445V Pt, H2(1 atm)/H* (test solution) // KCI(sat.)/Hg.Cl2/Hg. Calculate the pH of the unknown solution.

Consider the following cell: | Zn (s) | Ca2* (aq) , SO3²- (aq) || Hg²+ (aq) , F' (aq) | Pt (s) | al Indiate the Oxidizing Agent bindicate the ReducingAge Gont e Wnte he narreattions that occurs ai the cathode d) Write the half reactions that occurs at the anode e) Write the overall net cell reaction f) Calculate the cell potential g) Indicate if electricity is produced or consumed by the cell h) Indicate if the cell is galvanic or electrolytic

Consider the following cell reaction: Fe(s) + 2 H"(? М) —Ғе*"(1.00 М) + H-(g)(1.00 atm) →Fe? If the cell potential at 298 K is 0.164 volts, what is the pH of the hydrogen electrode? pH

The potential of the following voltaic cell is 0.55 V. Given this information, determinethe Kb for the organic base (RNH2) dissolved in the left cell.

Consider the following galvanic cell at 25°C. Pt | Cr²+ (0.25 M), Cr³+ (2.0 M) || Co²+ (0.17 M) | Co The overall reaction and equilibrium constant value are given below. 2 Cr²+ "(aq) + Co2+ Calculate the cell potential & for this galvanic cell and AG for the cell reaction at these conditions. + (aq) → 2 Cr³+ (aq) + Co(s) K = 2.79x107

e. Calculate the cell potential Ecell of the Galvanic cell (25°C). Give your answer in volts (V).

The condensed cell diagram for the reaction occurring in silver–zinc button batteries is  (-) Zn(s) | ZnO(s) | KOH(aq) || Ag2O(s) | Ag(s) (+)  Write the balanced equation for the net cell reaction equation

A voltaic cell is constructed in which the cathode is a standard hydrogen electrode and the anode is a hydrogen electrode ( PH, )= latm) immersed in a solution of unknown [H*]. If the cell potential is 0.201 V, what is the pH of the unknown solution at 298 K? pH =

An electrochemical cell consists of a silver metal electrode immersed in a solution with Ag* = 1.0 M separated by a porous disk from a copper metal electrode. If the copper electrode is placed in a solution of 5.0 MNH3 that is also 0.0080M in Cu(NH3),+, what is the cell potential at 25°C? Cu?+ (ag) + 4 NH3 (aq) = Cu(NH3)4²* (aq) K = 1.0 × 1013 Cell potential V