The 1876 United States presidential election in Iowa took place on November 7, 1876, as part of the 1876 United States presidential election. Voters chose 11 representatives, or electors, to the Electoral College, who voted for president and vice president.^[2]

1876 United States presidential election in Iowa

← 1872 November 7, 1876 1880 →
Turnout	24.57% 6.45 pp[1]
Nominee Rutherford B. Hayes Samuel J. Tilden Party Republican Democratic Home state Ohio New York Running mate William A. Wheeler Thomas A. Hendricks Electoral vote 11 0 Popular vote 171,326 112,121 Percentage 58.50% 38.28%
County Results Hayes   40-50%   50-60%   60-70%   70-80%   80-90%   90-100% Tilden   40-50%   50-60%   60-70%
President before election Ulysses S. Grant Republican Elected President Rutherford B. Hayes Republican

Iowa was won by Rutherford B. Hayes, the governor of Ohio (R-Ohio), running with Representative William A. Wheeler, with 58.50% of the vote, against Samuel J. Tilden, the former governor of New York (D–New York), running with Thomas A. Hendricks, the governor of Indiana and future vice president, with 38.28% of the popular vote.^[2]

The Greenback Party chose industrialist Peter Cooper and former representative Samuel Fenton Cary, received 3.22% of the vote. Iowa gave the Greenbacks both their largest number of total votes and percentage and Iowa would later be the last bastion of the congressional Greenbacks who held a House seat until 1889.