1920 Iowa gubernatorial election

Summary

The 1920 Iowa gubernatorial election was held on November 2, 1920. Republican nominee Nathan E. Kendall defeated Democratic nominee Clyde L. Herring with 58.66% of the vote.

1920 Iowa gubernatorial election

← 1918 November 2, 1920 1922 →
 
Nominee Nathan E. Kendall Clyde L. Herring
Party Republican Democratic
Popular vote 513,118 338,108
Percentage 58.66% 38.65%

Governor before election

William L. Harding
Republican

Elected Governor

Nathan E. Kendall
Republican

General election edit

Candidates edit

Major party candidates

  • Nathan E. Kendall, Republican
  • Clyde L. Herring, Democratic

Other candidates

  • George J. Peck, Socialist
  • Mathis Faber, Farmer–Labor
  • J. Jay Hisel, Socialist Labor

Results edit

1920 Iowa gubernatorial election[1]
Party Candidate Votes % ±%
Republican Nathan E. Kendall 513,118 58.66%
Democratic Clyde L. Herring 338,108 38.65%
Socialist George J. Peck 13,671 1.56%
Farmer–Labor Mathis Faber 9,153 1.05%
Socialist Labor J. Jay Hisel 760 0.09%
Majority 175,010
Turnout
Republican hold Swing

References edit

  1. ^ Kalb, Deborah (December 24, 2015). Guide to U.S. Elections. ISBN 9781483380353. Retrieved June 30, 2020.