1880 United States presidential election in Iowa

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.

1880 United States presidential election in Iowa

November 2, 1880
 
Nominee James A. Garfield Winfield S. Hancock James B. Weaver
Party Republican Democratic Greenback
Home state Ohio Pennsylvania Pennsylvania
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%

President before election

Rutherford B. Hayes
Republican

Elected President

James A. Garfield
Republican

Iowa voted for the Republican nominee, James A. Garfield, over the Democratic nominee, Winfield Scott Hancock. Garfield won the 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]

Results

1880 United States presidential election in Iowa[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican James Abram Garfield of Ohio Chester Alan Arthur of New York 183,904 56.99% 11 100.00%
Democratic Winfield Scott Hancock of Pennsylvania William Hayden English of Indiana 105,845 32.80% 0 0.00%
Greenback James Baird Weaver of Iowa Barzillai Jefferson Chambers of Texas 32,327 10.02% 0 0.00%
Prohibition Neal Dow of Maine Henry Adams Thompson of Ohio 592 0.18 0 0.00%
Total 322,668 100.00% 11 100.00%

Notes

    References

    1. "1880 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
    2. "1880 Presidential General Election Results - Iowa".
    This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.