1880 United States presidential election in Iowa

1880 United States presidential election in Iowa

← 1876 November 2, 1880 1884 →
 
Nominee James A. Garfield Winfield S. Hancock James B. Weaver
Party Republican Democratic Greenback
Home state Ohio Pennsylvania Iowa
Running mate Chester A. Arthur William H. English Barzillai J. Chambers
Electoral vote 11 0 0
Popular vote 183,904 105,845 32,327
Percentage 56.99% 32.80% 10.02%

County Results

President before election

Rutherford B. Hayes
Republican

Elected President

James A. Garfield
Republican

The 1880 United States presidential election in Iowa took place on November 2, 1880, as part of the 1880 United States presidential election. Voters chose 11 representatives, or electors to the Electoral College, who voted for president and vice president.

Iowa voted for the Republican nominee, James A. Garfield, over the Democratic nominee, Winfield Scott Hancock. Garfield won the state by a margin of 24.19%.

With 10.02% of the popular vote, Iowa would prove to be Greenback Party candidate James B. Weaver's second strongest state after Texas.[1]

  1. ^ "1880 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.

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